Problem: $h(t) = t^{2}-2t$ $g(x) = -6x^{2}-6x-7-5(h(x))$ $ g(h(1)) = {?} $
Solution: First, let's solve for the value of the inner function, $h(1)$ . Then we'll know what to plug into the outer function. $h(1) = 1^{2}+(-2)(1)$ $h(1) = -1$ Now we know that $h(1) = -1$ . Let's solve for $g(h(1))$ , which is $g(-1)$ $g(-1) = -6(-1)^{2}+(-6)(-1)-7-5(h(-1))$ To solve for the value of $g$ , we need to solve for the value of $h(-1)$ $h(-1) = (-1)^{2}+(-2)(-1)$ $h(-1) = 3$ That means $g(-1) = -6(-1)^{2}+(-6)(-1)-7+(-5)(3)$ $g(-1) = -22$